Let $h(x)=\begin{cases} 2^{x-1}&\text{for }x<1 \\\\ 2^{1-x}&\text{for }x\geq 1 \end{cases}$ Is $h$ continuous at $x=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Explanation: For $h$ to be continuous at $x=1$, we need $\lim_{x\to 1}h(x)$ and $h(1)$ to exist and be equal. Since $1\geq 1$, the rule that applies to $x=1$ is $2^{1-x}$. So $h(1)=2^{1-1}=1$. Now let's analyze $\lim_{x\to 1}h(x)$. Finding $\lim_{x\to 1^{ +}}h(x)$ For $x$ -values larger than $1$, the appropriate rule for $h(x)$ is $2^{1-x}$. Since $2^{1-x}$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 1^{ +}}h(x) \\\\ &=\lim_{x\to 1^{ +}}[2^{1-x}] \gray{2^{1-x}\text{ is the rule for }x>1} \\\\ &=2^{1-1} \gray{2^{1-x}\text{ is continuous at }x=1} \\\\ &=1 \end{aligned}$ Finding $\lim_{x\to 1^{ -}}h(x)$ For $x$ -values smaller than $1$, the appropriate rule for $h(x)$ is $2^{x-1}$. Since $2^{x-1}$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 1^{ -}}h(x) \\\\ &=\lim_{x\to 1^{ -}}[2^{x-1}] \gray{2^{x-1}\text{ is the rule for }x<1} \\\\ &=2^{1-1} \gray{2^{x-1}\text{ is continuous at }x=1} \\\\ &=1 \end{aligned}$ Conclusion We found that: $\lim_{x\to 1^{ +}}h(x)=\lim_{x\to 1^{ -}}h(x)=h(1)=1$ Since the one-sided limits are both equal to $h(1)$, we can determine that the two-sided limit $\lim_{x\to 1}h(x)$ is also equal to $h(1)$, and $h$ is continuous at $x=1$.